Pre-Algebra

10-3 Solving Equations with Variables on Both Sides10-3 Solving Equations with

Variables on Both Sides

Pre-Algebra

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Warm UpSolve.

1. 2x + 9x – 3x + 8 = 16

2. –4 = 6x + 22 – 4x

3. + = 5

4. – = 3

x = 1

x = -13

x = 34

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

27

x7 7

1

9x16

2x4

18

x = 50

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Problem of the Day

An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle?22.5 in.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Learn to solve equations with variables on both sides of the equal sign.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Some problems produce equations that have variables on both sides of the equal sign.

Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

A. 4x + 6 = x

Additional Example 1A: Solving Equations with Variables on Both Sides

4x + 6 = x– 4x – 4x

6 = –3x

Subtract 4x from both sides.

Divide both sides by –3.

–2 = x

6–3

–3x–3=

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

B. 9b – 6 = 5b + 18

Additional Example 1B: Solving Equations with Variables on Both Sides

9b – 6 = 5b + 18– 5b – 5b

4b – 6 = 18

4b 4

24 4 =

Subtract 5b from both sides.

Divide both sides by 4.

b = 6

+ 6 + 6

4b = 24Add 6 to both sides.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

C. 9w + 3 = 5w + 7 + 4w

Additional Example 1C: Solving Equations with Variables on Both Sides

9w + 3 = 5w + 7 + 4w

3 ≠ 7

9w + 3 = 9w + 7 Combine like terms.

– 9w – 9w Subtract 9w from both sides.

No solution. There is no number that can be substituted for the variable w to make the equation true.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

A. 5x + 8 = x

Try This: Example 1A

5x + 8 = x– 5x – 5x

8 = –4x

Subtract 4x from both sides.

Divide both sides by –4.

–2 = x

8–4

–4x–4=

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

B. 3b – 2 = 2b + 123b – 2 = 2b + 12

– 2b – 2b

b – 2 = 12

Subtract 2b from both sides.

+ 2 + 2

b = 14Add 2 to both sides.

Try This: Example 1B

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w

1 ≠ 8

3w + 1 = 3w + 8 Combine like terms.

– 3w – 3w Subtract 3w from both sides.

No solution. There is no number that can be substituted for the variable w to make the equation true.

Try This: Example 1C

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

To solve multistep equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

A. 10z – 15 – 4z = 8 – 2z - 15

Additional Example 2A: Solving Multistep Equations with Variables on Both Sides

10z – 15 – 4z = 8 – 2z – 15

+ 15 +15

6z – 15 = –2z – 7 Combine like terms.+ 2z + 2z Add 2z to both sides.

8z – 15 = – 7

8z = 8

z = 1

Add 15 to both sides.

Divide both sides by 8.8z 88 8=

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

B.

Additional Example 2B: Solving Multistep Equations with Variables on Both Sides

Multiply by the LCD.

4y + 12y – 15 = 20y – 14

16y – 15 = 20y – 14 Combine like terms.

y5

34

3y5

710

+ – = y –

y5

34

3y5

710

+ – = y –

20( ) = 20( )y5

34

3y5

710

+ – y –

20( ) + 20( ) – 20( )= 20(y) – 20( )y5

3y5

34

710

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 2B Continued

Add 14 to both sides.

–15 = 4y – 14

–1 = 4y

+ 14 + 14

–1 4

4y4 = Divide both sides by 4.

-14 = y

16y – 15 = 20y – 14

– 16y – 16y Subtract 16y from both sides.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Solve.

A. 12z – 12 – 4z = 6 – 2z + 32

Try This: Example 2A

12z – 12 – 4z = 6 – 2z + 32

+ 12 +12

8z – 12 = –2z + 38 Combine like terms.+ 2z + 2z Add 2z to both sides.

10z – 12 = + 38

10z = 50

z = 5

Add 12 to both sides.

Divide both sides by 10.10z 5010 10=

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

B.

Multiply by the LCD.

6y + 20y + 18 = 24y – 18

26y + 18 = 24y – 18 Combine like terms.

y4

34

5y6

68

+ + = y –

y4

34

5y6

68

+ + = y –

24( ) = 24( )y4

34

5y6

68

+ + y –

24( ) + 24( )+ 24( )= 24(y) – 24( )y4

5y6

34

68

Try This: Example 2B

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Subtract 18 from both sides.

2y + 18 = – 18

2y = –36

– 18 – 18

–36 2

2y2 = Divide both sides by 2.

y = –18

26y + 18 = 24y – 18

– 24y – 24y Subtract 24y from both sides.

Try This: Example 2B Continued

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3: Consumer Application

Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3 Continued

First solve for the price of one doughnut.

1.25 + 2d = 0.50 + 5dLet d represent the price of one doughnut.

– 2d – 2d

1.25 = 0.50 + 3dSubtract 2d from both sides.

– 0.50 – 0.50Subtract 0.50 from both sides.

0.75 = 3d

0.753

3d3= Divide both sides by 3.

0.25 = d The price of one doughnut is $0.25.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Additional Example 3 Continued

Now find the amount of money Jamie spends each morning.

1.25 + 2d Choose one of the original expressions.

Jamie spends $1.75 each morning.

1.25 + 2(0.25) = 1.75

0.25n0.25

1.75 0.25 =

Let n represent the number of doughnuts.

Find the number of doughnuts Jamie buys on Tuesday.

0.25n = 1.75

n = 7; Jamie bought 7 doughnuts on Tuesday.

Divide both sides by 0.25.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Try This: Example 3

Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Try This: Example 3 Continued

First solve for distance around the track.

2x + 4 = 4x + 2Let x represent the distance around the track.

– 2x – 2x

4 = 2x + 2Subtract 2x from both sides.

– 2 – 2 Subtract 2 from both sides.

2 = 2x

22

2x2= Divide both sides by 2.

1 = x The track is 1 mile around.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Try This: Example 3 Continued

Now find the total distance Helene walks each day.

2x + 4 Choose one of the original expressions.

Helene walks 6 miles each day.2(1) + 4 = 6

Let n represent the number of 1-mile laps.

Find the number of laps Helene walks on Saturdays.

1n = 6

Helene walks 6 laps on Saturdays.

n = 6

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Journal: Describe the kind of solution that has no solution.

Pre-Algebra

10-3 Solving Equations with Variables on Both Sides

Lesson Quiz

Solve.

1. 4x + 16 = 2x

2. 8x – 3 = 15 + 5x

3. 2(3x + 11) = 6x + 4

4. x = x – 9

5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each?

x = 6

x = –8

no solution

x = 3614

12

An orange has 45 calories. An apple has 75 calories.